Question

Equation of a straight line passing through the point of intersection of $x-y+1=0$ and $3x+y-5=0$ and perpendicular to one of them is

• A. $x+y+3=0$
• B. $x+y-3=0$
• C. $x-3y-5=0$
• D. $x-3y+5=0$

Answer 1

Answer: (B), (D)

$x-3y+5=0$
$x+y-3=0$

Given Lines

$x-y+1=0$------(1)

$3x+y-5=0$------(2)

from eq (1)

$y=x+1$

putting y in eq (2)

$3x+x+1-5=0$

$4x=4$

$x=1$

then $y=2$

Point of intersection be P$(1,2)$

now, slope of new lines $\perp$ to given line would be $m_1=-1$

(as it is -ve reciprocal)

$m_2=\frac{1}{3}$

Hence eq are

$x+y=3$ ($m_1=-1$)

and

$x-3y+5=0$ ($m_2=\frac{1}{3}$)